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Asymptotic Theory for Adaptive Allocation · Tutorial 207 of 501

Consistency of Arm-Specific Means Under Adaptive Sampling

Learn why an arm-specific mean can converge under outcome-adaptive assignment, which moment and sampling conditions suffice, and how to diagnose failures.

Advanced 10 min read

What You'll Learn

  • State sufficient conditional-mean, moment, and persistent-sampling assumptions for arm-specific consistency.
  • Prove consistency by indexing responses in order within each arm and applying a martingale argument.
  • Distinguish infinitely many assignments from a positive limiting allocation fraction.
  • Check how sparse adaptive sampling affects consistency without implying a central limit theorem.
  • Identify failures caused by finite sampling or changing conditional response means.

Why Adaptive Sampling Need Not Prevent Consistency

In a response-adaptive trial, the probability of assigning a participant to a treatment can depend on responses observed earlier. The resulting allocation count is random, and it may be associated with the observed history. That dependence does not, by itself, invalidate the sample mean within a treatment. The key distinction is between how often a treatment is selected and the conditional distribution of its response when selected.

This tutorial gives a sufficient condition for the mean of the responses observed on each arm to converge to that arm's response mean. The result is an almost-sure consistency statement: with probability one, the estimation error tends to zero as the number of assignments grows. It does not require an arm's fraction of all assignments to settle at a positive value. It does require that the arm continue to be sampled, and that its conditional response mean remain stable.

Setup and Assumptions

Consider a sequential trial with a fixed finite number \(K\) of treatments. At assignment \(t\), let \(A_t\in\{1,\ldots,K\}\) denote the selected treatment, and let \(Y_t\) be the response observed after that assignment. For arm \(i\), write \(I_{i,t}=1\) if \(A_t=i\), and \(I_{i,t}=0\) otherwise. The number assigned to arm \(i\) by time \(n\) and its observed sample mean are

$$ N_i(n)=\sum_{t=1}^n I_{i,t}, \qquad \bar Y_i(n)=\frac{\sum_{t=1}^n I_{i,t}Y_t}{N_i(n)} \quad\text{when }N_i(n)>0. $$

Let \(\mathcal H_{i,k-1}\) denote the information available immediately before the response from the \(k\)th assignment to arm \(i\) is generated. This includes the past trial history and the fact that arm \(i\) has just been selected; it does not include that response. Write \(Y_{i,k}\) for this \(k\)th observed response on arm \(i\). A sufficient set of assumptions is:

  • Stable conditional mean: for every \(k\), \(E(Y_{i,k}\mid\mathcal H_{i,k-1})=\mu_i\), where \(\mu_i\) is a fixed, finite arm-specific mean.
  • Bounded conditional second moment: for some finite constant \(C_i\), \(E\{(Y_{i,k}-\mu_i)^2\mid\mathcal H_{i,k-1}\}\le C_i\) for every \(k\).
  • Persistent sampling: \(N_i(n)\to\infty\) almost surely as \(n\to\infty\).

The response assumption says that, conditional on the information available before an arm-\(i\) response, its expected value is \(\mu_i\). Its conditional distribution may otherwise depend on the past. In the familiar special case where each arm has an independent, identically distributed response distribution with finite variance and mean \(\mu_i\), the first two conditions hold. The more general conditional formulation makes explicit what is needed when the trial is adaptive.

Theorem: Under the stable conditional mean and bounded conditional second moment assumptions, if arm \(i\) is sampled infinitely often almost surely, then \(\bar Y_i(n)\to\mu_i\) almost surely. The conclusion holds for each arm satisfying these conditions.

Why the Result Holds

The essential step is to index the observations by their order within arm \(i\), rather than by calendar assignment time. Define the centered response \(D_{i,k}=Y_{i,k}-\mu_i\). By the conditional-mean assumption, these centered responses have conditional expectation zero given the information before their respective responses. They are therefore martingale differences with respect to the history observed at successive arm-\(i\) responses. Their conditional second moments are bounded by \(C_i\).

For clarity, fix one arm and omit its subscript. Consider the weighted partial sums \(\sum_{k=1}^m D_k/k\). The martingale differences \(D_k/k\) have a finite total sum of expected conditional variances:

$$ \sum_{k=1}^{\infty} E\left[\left.\frac{D_k^2}{k^2}\right|\mathcal H_{k-1}\right] \le C_i\sum_{k=1}^{\infty}\frac{1}{k^2}<\infty. $$

The martingale convergence theorem consequently gives convergence almost surely of \(\sum_{k=1}^m D_k/k\) as \(m\to\infty\). Kronecker's lemma states that if a series \(\sum_{k=1}^{\infty}x_k/k\) converges, then \(m^{-1}\sum_{k=1}^m x_k\to0\). Applying it to \(x_k=D_k\) yields

$$ \frac{1}{m}\sum_{k=1}^m D_k = \frac{1}{m}\sum_{k=1}^m(Y_{i,k}-\mu_i) \longrightarrow 0 \quad\text{almost surely}. $$

At calendar time \(n\), the arm-specific mean is exactly the mean of the first \(N_i(n)\) responses indexed within that arm. Persistent sampling says this random index tends to infinity almost surely. Substituting \(m=N_i(n)\) into the within-arm convergence therefore gives

$$ \bar Y_i(n)-\mu_i = \frac{1}{N_i(n)} \sum_{k=1}^{N_i(n)}(Y_{i,k}-\mu_i) \longrightarrow 0 \quad\text{almost surely}. $$

This proof explains why adaptive assignment does not automatically spoil consistency. Assignment decisions can depend on past responses, but the next response on an arm still has the specified conditional mean, and enough responses from that arm eventually accumulate. The denominator is random, yet it is simply the number of terms in the within-arm average.

Worked Example: Calculating Arm-Specific Means in an Adaptive Trial

Suppose a hypothetical two-arm trial assigns participants sequentially. After each response, the allocation rule may use the accumulated trial history to choose the next arm. The first seven observed assignments and responses are \((A,7),(B,4),(A,5),(A,8),(B,6),(A,6),(B,5)\), where each pair records the assigned arm and response in response units.

For arm A, the count is \(N_A(7)=4\), and its observed responses are \(7,5,8,6\). Thus

$$ \bar Y_A(7)=\frac{7+5+8+6}{4}=\frac{26}{4}=6.50 \quad\text{response units}. $$

For arm B, the count is \(N_B(7)=3\), and its responses are \(4,6,5\). Therefore

$$ \bar Y_B(7)=\frac{4+6+5}{3}=\frac{15}{3}=5.00 \quad\text{response units}. $$

These are the observed arm-specific means at assignment 7. To apply the consistency theorem as \(n\) grows, one must additionally assume stable conditional response means and bounded conditional second moments for both arms, and that both arms are sampled infinitely often. Seven observations alone cannot establish those long-run assumptions or prove that either estimate is close to its target.

Infinite Sampling Is Not the Same as a Positive Allocation Fraction

Persistent sampling requires \(N_i(n)\to\infty\); it does not require \(N_i(n)/n\) to converge to a positive constant. An arm may receive an increasingly small share of assignments and still be sampled infinitely many times. Its mean can then be consistent, although this theorem gives no guarantee that convergence is fast or that a usual standard-error approximation is accurate.

Worked Example: A Vanishing Assignment Probability

Consider a hypothetical two-arm design in which arm A is assigned at time \(t\) with conditional probability \(p_t\). Let \(R_{t-1}\) be an indicator determined by past outcomes, and set \(p_t=(1+R_{t-1})/(t+2)\). Thus \(p_t\) adapts to the observed history, while

$$ \frac{1}{t+2}\le p_t\le\frac{2}{t+2}. $$

The lower bound implies \(\sum_{t=1}^{\infty}p_t=\infty\). If the assignment is randomized with this conditional probability at each time, the conditional Borel–Cantelli result implies that arm A is assigned infinitely often almost surely. In particular, its sample count diverges even though \(p_t\le 2/(t+2)\to0\). The long-run average assignment probability up to \(n\) is bounded above by

$$ \frac{1}{n}\sum_{t=1}^n p_t \le \frac{2}{n}\sum_{t=1}^n\frac{1}{t+2} \longrightarrow 0. $$

If the conditional response mean on arm A is a fixed \(\mu_A\) and its conditional variance is bounded, the theorem gives \(\bar Y_A(n)\to\mu_A\) almost surely. This conclusion is about consistency, not speed: the arm receives a vanishing share of the trial, so its estimate may stabilize slowly. The allocation rule must also ensure that the lower bound on \(p_t\) really holds; adaptation alone does not guarantee persistent sampling.

Failure Modes and Diagnostics

The theorem is a sufficient result, not an automatic property of every adaptive trial. Its assumptions clarify what can go wrong and what should be checked before interpreting an arm-specific average as an estimate of a stable mean.

  • Sampling stops or is too sparse. If the assignment rule can permanently exclude an arm, its count may remain finite, so its sample mean cannot be justified by this result. A design-level lower bound on assignment probabilities, or a direct argument that the count diverges, is more informative than observing that the arm was assigned several times early in a trial.
  • The conditional mean changes. If later responses on an arm have a different conditional mean because of time trends, changing eligibility, or a response process altered by prior treatment, then there may be no fixed \(\mu_i\) to which the raw mean should converge. Check whether the response model makes the conditional mean stable for the actual sequence of participants and trial times.
  • Conditional moments are too large. A bounded conditional second moment is a convenient sufficient condition. Extremely heavy-tailed responses, or variance that grows without control along the sequence, fall outside the stated proof. A few finite-sample summaries do not verify this assumption; it must be justified from the response model or supported by substantive knowledge.
  • The response model changes under selection. The assumption concerns the response conditional on the history and the arm selection. If selection changes which response distribution is being sampled in a way that changes its conditional mean, the observed arm mean may target a changing mixture rather than one fixed arm mean.

A diagnostic review should therefore examine the allocation mechanism and the response model together. Plotting \(N_i(n)\) over time can reveal an arm whose accrual is slowing or stopping, but a plot cannot establish infinite sampling. Monitoring response averages can reveal drift, but apparent stability in a finite trial does not prove that the conditional mean is constant. The assumptions are primarily design and model conditions, not tests that can be conclusively certified from the observed data.

Worked Example: When an Arm May Be Sampled Only Finitely Often

Suppose a hypothetical design assigns a participant to arm A at time \(t\) with probability \(p_t=2^{-t}\), independently at each assignment. The expected total number of assignments to A is

$$ E\{N_A(\infty)\} =\sum_{t=1}^{\infty}2^{-t} =1. $$

Because \(N_A(\infty)\) is a nonnegative integer-valued random variable with finite expectation, it must be finite almost surely; an infinite value with positive probability would make its expectation infinite. Thus the persistent-sampling assumption fails. If, for example, A is assigned twice and its two hypothetical responses are 2 and 4 response units, the observed mean is \((2+4)/2=3.00\) response units. That calculation is valid for the two observations, but the design does not provide an increasing sequence of arm-A observations to which the consistency theorem can be applied. The issue is not the arithmetic; it is the finite total sample count.

How This Result Differs from Related Methods

Under fixed allocation, the ordinary strong law of large numbers gives consistency for an arm mean when responses on that arm are identically distributed with finite mean (and suitable independence conditions). The adaptive result above uses a martingale strong-law argument instead: it permits dependence through the past, provided each response has the same conditional mean and the moment condition holds. In both settings, infinitely many observations from the arm are essential to the stated argument.

An inverse-probability-weighted estimator is a different construction: it reweights observations using their assignment probabilities and can target a population quantity under appropriate conditions. It is not needed merely to prove consistency of the raw within-arm sample mean under the conditional response assumptions stated here. Conversely, if those assumptions fail, this theorem does not establish that reweighting will repair the problem; that requires its own estimand and assumptions.

Consistency also does not specify the distribution of estimation error at a particular rate. A count that diverges very slowly can still yield almost-sure convergence while leaving substantial uncertainty at a finite sample size. A central limit theorem requires additional conditions and is a separate conclusion; it should not be inferred from consistency alone. Likewise, consistency of arm-specific means says nothing by itself about convergence of allocation proportions or the validity of a causal interpretation without an appropriate trial design and estimand.

Key takeaway: For an arm-specific sample mean to converge under adaptive assignment, a sufficient route is a stable conditional response mean, controlled conditional second moments, and infinitely many assignments to that arm. A positive limiting allocation fraction is not required.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions below.

  1. State the three assumptions in the consistency theorem and identify which one concerns the allocation rule.
  2. Arm C is sampled infinitely often, has a fixed conditional mean of 12 milligrams, and satisfies the bounded conditional second-moment condition. What does the theorem imply about its sample mean?
  3. Why does a positive limiting allocation fraction suffice for persistent sampling, but persistent sampling not require a positive limiting fraction?
  4. Suppose an arm's conditional mean is 5 units for its first 100 responses and 10 units thereafter. Which assumption needed for the theorem fails?

Solutions. (1) The assumptions are a fixed conditional mean, a uniformly bounded conditional second moment, and an arm count that diverges almost surely. Persistent sampling is the condition directly about the allocation rule. (2) Its sample mean converges almost surely to 12 milligrams. (3) If \(N_i(n)/n\to q_i>0\), then \(N_i(n)\to\infty\), since \(N_i(n)\) is asymptotic to \(q_i n\). The reverse implication does not hold: a count can diverge while its fraction of \(n\) tends to zero. (4) The stable conditional-mean assumption fails because there is no single fixed conditional mean for the whole sequence of responses.

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